# Question Video: Finding the First Derivative of an Exponential Function

Use the integral test to determine whether the series ∑_(𝑛 = 1)^(∞) (1/𝑛) converges or diverges.

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### Video Transcript

Use the integral test to determine whether the series, which is the sum from 𝑛 equals one to ∞ of one over 𝑛, converges or diverges.

Let’s start by recalling the integral test. The integral test tells us that if 𝑓 is a continuous, positive, decreasing function on the interval between 𝑘 and ∞ and that 𝑓 of 𝑛 is equal to 𝑎 𝑛. Then, if the integral from 𝑘 to ∞ of 𝑓 of 𝑥 with respect to 𝑥 is convergent, so is the sum from 𝑛 equals 𝑘 to ∞ of 𝑎 𝑛. And secondly, if the integral from 𝑘 to ∞ of 𝑓 of 𝑥 with respect to 𝑥 is divergent, so is the sum from 𝑛 equals 𝑘 to ∞ of 𝑎 𝑛. Now, in our case, our series is the sum from 𝑛 equals one to ∞ of one over 𝑛. Therefore, 𝑎 𝑛 is equal to one over 𝑛. Using the fact that 𝑓 of 𝑛 is equal to 𝑎 𝑛, we can say that 𝑓 of 𝑥 is equal to one over 𝑥.

Now, we need to check that 𝑓 of 𝑥 is a continuous, positive, and decreasing function on the interval between 𝑘 and ∞. Noting that since our sum goes from 𝑛 equals one, 𝑘 must be equal to one. Let’s start by checking the continuity of our function over this interval. The only discontinuity of 𝑓 of 𝑥 occurs when 𝑥 is equal to zero. However, zero is not in our interval between one and ∞. Therefore, this function must be continuous between one and ∞. On this interval, 𝑥 is always positive. Therefore, one over 𝑥 must also be positive. And so, we’ve satisfied that condition too. Now, if 𝑥 starts at one and gets larger and larger and larger, then one over 𝑥 will get smaller and smaller and smaller. Therefore, we can say that this is a decreasing function on our interval.

Now that we’ve satisfied these three conditions, we’re able to use the integral test. We need to work out whether the integral from one to ∞ of one over 𝑥 with respect to 𝑥 is convergent or divergent. Now, we know that the differential of the natural logarithm of 𝑥 with respect to 𝑥 is equal to one over 𝑥. Therefore, the natural logarithm of 𝑥 is the antiderivative of one over 𝑥. Hence, we have that our integral is equal to the natural logarithm of 𝑥 between one and ∞. When using the bound of ∞, we need to take the limit as 𝑥 tends to ∞ of the natural logarithm of 𝑥. Then, for our lower bound of one, we simply subtract the natural logarithm of one.

Now, let’s evaluate this limit, the natural logarithm as an increasing function. Therefore, as 𝑥 gets larger, the natural logarithm of 𝑥 also gets larger. Hence, we can say that this limit is also equal to ∞. We also have that the natural logarithm of one is equal to zero. However, regardless of what constant this term is, this will not affect our answer of ∞. This tells us that our integral must be divergent. Hence, by the integral test, we can say that the sum from 𝑛 equals one to ∞ of one over 𝑛 is divergent.